Question

For the reaction, \[\ce{A2(g) + B2(g) <=> 2AB(g); \Delta H}\] is -ve.

the following molecular scenes represent differenr reaction mixture. (A-green, B-blue)

Closed  ←
System	At equilibrium	(x)	(y)

1. Calculate the equilibrium constant K_p and (K_c).
2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
3. What is the effect of an increase in pressure for the mixture at equilibrium?

Answer 1

K_c = `(["AB"]^2)/(["A"_2]["B"_2])`; A – green, B – blue

Given that ‘V’ is constant(closed system)

At equilbrium,

K_c = `((-))/((2/"V")(2/"V")) = 16/4 = 4`

K_p = K_c(RT)^∆n = 4 (RT)⁰ = 4

At stage ‘x’,

Q = `((6/"V")^2)/((2/"V")(1/"V")) = 36/2 = 18`

Q > K_c i.e., reverse reaction is favoured.

At stage ‘y’,

Q = `((3/"V")^2)/((3/"V")(3/"V")) = 9/(3 xx 3) = 1`

K_c > Q i.e., forward reaction is favoured.